Can you prove this trigonometric inequality?

Question

In acute-angled triangle ABC, prove that [CosA/CosB]² + [CosB/CosC]² +[CosC/CosA]² + 8 CosA CosB CosC ≥ 4.
This was originally asked by Annie G without the condition of the triangle being acute-angled and gianlino had shown that the inequality does not hold for all triangles giving an example taking angles of the triangle as 30°, 30° and 120°. He informed me that the inequality is indeed true with the condition that the triangle is acute-angled, but I do not know the proof.

gianlino · Accepted Answer

For x real x^2 >= 2x - 1 and for y real >0 

(2y - 1)^2 >= 0  ===>  4y^2 - 4y + 1 >= 0  ===> 1/y > 4 - 4y.

Combining both we get 

[CosA/CosB]^2 >= 2(cos A / cos B) - 1 >= 2 cos(A) (4 - 4 cosB) - 1 

= 8cos A - 8 cos A cos B - 1.

Adding the three corresponding terms you want

8 (cos A + cos B + cos C) - 8 (cos A cos B + cos C cos B + cos A cos C) - 3 

+ 8 cos A cos B cos C >= 4

This can be rewritten as   1 >= 8 (1- cos A)(1 - cos B)(1- cos C)

or 1/64 >= (sin(A/2) sin(B/2) sin(C/2))^2

This in turn follows from the inequality sin x sin y <= (sin((x+y)/2))^2. This inequality implies that the max of (sin(x) sin(y) sin(z))^2 for x,y,z in [0,pi/2] and x + y + z = pi/2 is attained for x = y = z = pi / 6

and equals  1/2^6 = 1/64.

sri · Answer

Nice one. My gut feel was to start with Jenson's inequality,
A+B+C=π then cosA+cosB+cosC ≤ 3/2
I went in a couple of seemingly promising ways, but did not reach critical mass (I guess you probably may have tried something on those same lines, Madhukarji).
Will try again and edit with the solution, hopefully before someone else gets it.

Whoa, too late.. Looks like gianlino has come up with the solution.  Maybe he deserves the BA.

bskelkar · Answer

I think at the first look it needs AM ≥ GM 
LHS = [(cosA/cosB)+(cosB/cosC)+cosC/cosA)]^2 - 6 + 8cosAcosBcosC
≥9 - 6 + 8cosAcosBcosC 
But For want of time I will come back to this and solve it.

sogol · Answer

This is a shortcut. Meanwhile i'll try to feel of a more rational manner. Put A = B = C = 60 on all sides of the eq LHS is 1/2*half*1/2 = 1/8 RHS is 1/eight LHS = RHS now put A = B = 60 and C = 90 LHS = 1/4 RHS = 0 LHS > RHS consequently LHS >=RHS

ajay · Answer

hii please can anyone solve this:-
a car A starts from a point P towards another point Q.Another car B starts (also from p) 1 h after the first car and overtakes it after covering 30% of the distance PQ .after that the cars continue and on reaching Q, Car B reverses and meet car A at 23(1/3) of the distance QP.find the time taken by car B to cover the distance PQ(in hours)

Anonymous · Answer

Yeah, it's a tricky one. I couldn't figure it out but I have some links with the proof. This question is the same as question 50 in 

http://adria.inaoep.mx/~diplomados/biblio/GPyT/103ProblemasTrigonometry/5solutionstoAdvancedproblems.pdf

The proof there also cites results from 

http://adria.inaoep.mx/~diplomados/biblio/GPyT/103ProblemasTrigonometry/4SolutionstoInroductoryproblems.pdf

Can you prove this trigonometric inequality?

6 Answers